How do you find the critical numbers for #g(t) = t sqrt(4-t) # to determine the maximum and minimum?

1 Answer
Aug 3, 2018

Answer:

#PMAX(8/3;(16*sqrt(3))/9)#

Explanation:

Given #g(t)=tsqrt(4-t)#
we get by the product and the chain rule

#g'(t)=sqrt(4-t)+t*(1/2)(4-t)^(-1/2)*(-1)#
which simplifies to

#g'(t)=sqrt(4-t)-t/(2sqrt(4-t))#
so we get

#g*(t)=0# if

#sqrt(4-t)=t/(2sqrt(4-t))# Multiplying with #2sqrt(4-t)#

#2(4-t)=t#

so #t=8/3#

#g''(t)=1/2(4-t)^(-1/2) * (-1)-1/2*(4-t)^(-1/2)+t/2(4-t)^(-3/2) * (-1/2)#

#g''(t)=t/sqrt(4-t)-t/(4(4-t)^(3/2))#

#g''(8/3)=-3sqrt(3)/4<0#
so we get #PMAX(8/3;16*sqrt(3)/9)#